At time , a disk of radius starts to roll without slipping on a horizontal plane with an angular acceleration of . A small stone is stuck to the disk. At , it is at the contact point of the disk and the plane. Later, at time , the stone detaches itself a

Question

At time $t = 0$ , a disk of radius $1 m$ starts to roll without slipping on a horizontal plane with an angular acceleration of $α = \frac{2}{3} {rad s}^{- 2}$ . A small stone is stuck to the disk. At $t = 0$ , it is at the contact point of the disk and the plane. Later, at time $t = \sqrt{π} s$ , the stone detaches itself and flies off tangentially from the disk. The maximum height (in m) reached by the stone measured from the plane is $\frac{1}{2} + \frac{x}{10}$ . The value of $x$ is _____. [Take $g = 10 {ms}^{- 2}$ ] [2022]

Smart Achievers · Accepted Answer

(0.52)

The angle rotated by disc in $t = \sqrt{π} sec is$

$θ = ω_{0} t + \frac{1}{2} α t^{2} = \frac{1}{2} \times \frac{2}{3} {(\sqrt{π})}^{2} = \frac{π}{3} rad$

Angular velocity at the end of $t = \sqrt{π} sec is$

$ω = ω_{0} + α t = \frac{2}{3} \sqrt{π} rad/s$

and, as it is pure rolling so $V_{c m} = ω R = \frac{2}{3} \sqrt{π} \times 1 m/s$

So, at the moment of detachment we get the following situation

So, $V = \sqrt{V_{c m}^{2} + {(ω R)}^{2} + 2 (ω R) (V_{c m}) \cos 120 °}$

$= V_{c m}$ $[∵ V_{c m} = ω R]$

$= \frac{2 \sqrt{π}}{3} m/s$

$\tan θ = \frac{(ω R) \sin 120 °}{V_{c m} + (ω R) \cos 120 °} = \frac{\frac{\sqrt{3}}{2}}{1 - \frac{1}{2}} = \sqrt{3}$

$∴$ $θ = 60 °$

$Then, H_{\max} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{{(\frac{2 \sqrt{π}}{3})}^{2} \times \frac{3}{4}}{2 \times 10} = \frac{\frac{4 π}{9} \times \frac{3}{4}}{20} = \frac{π}{60} m$

$Height from ground = R (1 - \cos \frac{π}{3}) + \frac{π}{60} = \frac{R}{2} + \frac{π}{60} = \frac{1}{2} + \frac{π}{60} = \frac{1}{2} + \frac{1}{10} (\frac{π}{6})$

$∴$ $x = \frac{π}{6} \approx 0.52$

Solution

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